How do you do this

[mollymvora]

How many license plates of 4 symbols can be made using two letters and 2 digits. I got 540,800. is that correct? Also, what about when it says how many multiples of 3 less than 1000 can be formed from digits 2,5, and 9. is it 10. Is there any quick way to figure the second problem out? I had to write all the 3 digit ones out and see if they add up to a multiple of three, so is there any faster way?

How do you know how many possibilities you can get?
Like for #1 I know it is some combination of 26 * 26* 10 *10, but how do I calculate all the different ways that the numbers and letters can be arranged?

 

[Deleted member 4993]

How many license plates of 4 symbols can be made using two letters and 2 digits. I got 540,800. is that correct?

Also, what about when it says how many multiples of 3 less than 1000 can be formed from digits 2,5, and 9. is it 10. Is there any quick way to figure the second problem out?
For a number to be divisible by 3 - the digits must add up to 9, 6 or 3

so the digits could be

222
555
999
255
525
552

I don't see any more!

What other numbers did you get?

I had to write all the 3 digit ones out and see if they add up to a multiple of three, so is there any faster way?

How do you know how many possibilities you can get?
Like for #1 I know it is some combination of 26 * 26* 10 *10, but how do I calculate all the different ways that the numbers and letters can be arranged?

 

[soroban]

Hello, mollymvora!

How many multiples of 3 less than 1000 can be formed from digits {2, 5, 9} ?

Subhotosh is correct . . . basically.

The sum of the digits will be a multiple of 3.
So the sum can be: 3, 6, 9, 12, 15, 18, 21, 24, or 27.


I found eight such numbers:

. . \(\displaystyle \begin{array}{c} 222 \\ 555\\ 999 \end{array} \quad \begin{array}{c} 225 \\ 252 \\ 522 \end{array} \quad \begin{array}{c}9 \\ 99 \end{array}\)

 

[Deleted member 4993]

Soroban is absolutely right - in my haste, I had incorrectly assumed that the numbers had to be 3 digit numbers.

Denis - I am going ... I am going .... right to the corner....

 

[soroban]

Hello again, mollymvora!

The problem is not stated clearly.

How many license plates of 4 symbols can be made using two letters and 2 digits.
I got 540,800. is that correct? . No

\(\displaystyle \text{[A] Suppose it means: two letters, then two digits . . . }in\:that\rder.\)

. . \(\displaystyle \text{Then there are: }\:26\cdot26\cdot10\cdot10 \:=\:67,600\text{ possible license plates.}\)


\(\displaystyle \text{ Suppose it means: two letters and two digits . . . }in\:any\rder.\)

\(\displaystyle \text{There are four cases to consider . . .}\)

\(\displaystyle \text{(1) Two different letters, two different digits: }ABxy\)
. . .\(\displaystyle \text{There are: }\,26\cdot25\cdot10\cdot9 \,=\,58,\!500\text{ selections and }4! \,=\,24\text{ permutations.}\)
. . .\(\displaystyle \text{There are: }58,\!500 \times 24 \:=\:1,404,000\text{ such plates.}\)

\(\displaystyle \text{(2) Two identical letters, two different digits: }AAxy\)
. . .\(\displaystyle \text{There are: }\,26\cdot10\cdot9 = 2,\!340\text{ selections and }\tfrac{4!}{2!} = 12\text{ permutations.}\)
. . .\(\displaystyle \text{There are: }2,\!340 \times 12 \:=\:28,\!080\text{ such plates.}\)

\(\displaystyle \text{(3) Two different letters, two identical digits: }ABxx\)
. . .\(\displaystyle \text{There are: }\;26\cdot25\cdot10 = 6,\!500\text{ selections and }\tfrac{4!}{2!} = 12\text{ permutations.}\)
. . .\(\displaystyle \text{There are: }6,\!500 \times 12 \:=\:78,\!000\text{ such plates.}\)

\(\displaystyle \text{(4) Two identical letters, two identical digits: }AAxx\)
. . .\(\displaystyle \text{There are: }\,26\cdot10 = 260\text{ selections and }\tfrac{4!}{2!2!} = 6\text{ permutations.}\)
. . .\(\displaystyle \text{There are: }260 \times 6 \:=\:1,\!560\text{ such plates.}\)

\(\displaystyle \text{Therefore, there are: }\,1,\!404,\!000\,+\,28,\!080\,+\,78,\!000\,+\,1,\!560 \;=\;1,\!511,\!640\text{ possible license plates.}\)